Question: Let $R$ be the region enclosed by $y=3-x^2$ and the axes in the first quadrant. $y$ $x$ $(0,3)$ $(\sqrt 3,0)$ $ R$ ${y=3-x^2}$ Region $R$ is the base of a solid whose cross sections perpendicular to the $y$ -axis are squares. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{0}^{3} (3-y)\,dy$ (Choice B) B $\int_{0}^{\sqrt 3} (3-y)\,dy$ (Choice C) C $\int_{0}^{\sqrt 3} (3-y)^2\,dy$ (Choice D) D $\int_{0}^{3} (3-y)^2\,dy$
Let's imagine the solid is made out of many thin slices that are perpendicular to the $y$ -axis. $y$ $x$ $(0,3)$ $(\sqrt 3,0)$ Each slice is a prism. Let the width of each slice be $dy$ and let the area of the prism's face, as a function of $y$, be $A(y)$. Then, the volume of each slice is $A(y)\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(y)\,dy$ What we now need is to figure out the expression of $A(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ $(0,3)$ $(\sqrt 3,0)$ ${y=3-x^2}$ $ s(y)$ $ s(y)$ $ dy$ $ A(y)$ The face of that slice is a square with side $s(y)$, which is equal to the distance from the curve $y=3-x^2$ to the $y$ -axis. To find an expression for $s(y)$, we must rewrite $y=3-x^2$ as a function of $y$ (which means we need to isolate $x$ ). $x=\sqrt{3-y}$ So, for each $y$ -value, this is the side of the square: $s(y)=\sqrt{3-y}$ Now we can find the area of the square: $\begin{aligned} &\phantom{=}A(y) \\\\ &=[s(y)]^2 \\\\ &=(\sqrt{3-y})^2 \\\\ &=3-y \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=3$. So the interval of integration is $[0,3]$. Now we can express the definite integral in its entirety! $\int_0^3 (3-y)\,dy$